1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differantiation proof question exlanation problem

  1. Jan 21, 2009 #1
    if f(x) is differentiable on x_0
    prove that
    [​IMG]

    ??

    [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/itex]

    now let [itex] u = x+h \Rightarrow x = u-h [/itex]

    So [itex] f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h} [/itex]

    Since [itex] h \to 0, f'(u-h) \to f'(u) [/itex]

    Therefore [itex] f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h} [/itex]

    [itex] f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)[/itex]

    [itex] 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h} [/itex]

    Hence, [itex] f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} [/itex]


    i understand this solution but
    how to pick those variables??
    and why it didnt use the point x_0
    that they presented
     
  2. jcsd
  3. Jan 21, 2009 #2
    there is another definition to limit that uses x_0 that i was given
    [itex]
    f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}
    [/itex]

    is there a way to use this definition like you did before
    what variables to pick so it will go on the "h" defnition??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Differantiation proof question exlanation problem
Loading...